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Joe makes a one-time deposit of \[tex]$5,000.00 to open an interest-bearing savings account for his newborn grandson, Tyler. The table shows the expected balance of the account, \(B(t)\), when Tyler is \(t\) years old.

\[
\begin{tabular}{|c|c|}
\hline
Age in years, \(t\) & Balance in dollars, \(B(t)\) \\
\hline
0 & \$[/tex]5,000.00 \\
\hline
2 & \[tex]$5,030.08 \\
\hline
4 & \$[/tex]5,060.34 \\
\hline
6 & \[tex]$5,090.78 \\
\hline
8 & \$[/tex]5,121.41 \\
\hline
10 & \[tex]$5,152.21 \\
\hline
12 & \$[/tex]5,183.21 \\
\hline
14 & \$5,214.39 \\
\hline
\end{tabular}
\]

Consider five intervals representing periods of [tex]\(t\)[/tex] years during which the account balance, [tex]\(B(t)\)[/tex], changed. Order the intervals from greatest to least average rate of change.

[tex]\[
[10,14] \quad [0,4] \quad [0,14] \quad [6,12] \quad [2,8]
\][/tex]



Answer :

GinnyAnswer
To solve this problem, we need to determine the average rate of change of the balance for each of the given time intervals and then order them from greatest to least.

The average rate of change between two points [tex]\((t_1, B(t_1))\)[/tex] and [tex]\((t_2, B(t_2))\)[/tex] is calculated as:

[tex]\[ \text{Average Rate of Change} = \frac{B(t_2) - B(t_1)}{t_2 - t_1} \][/tex]

We will calculate this for each interval:

### Interval [10, 14]
- [tex]\(t_1 = 10\)[/tex], [tex]\(B(t_1) = 5152.21\)[/tex]
- [tex]\(t_2 = 14\)[/tex], [tex]\(B(t_2) = 5214.39\)[/tex]

[tex]\[ \text{Average Rate of Change} = \frac{5214.39 - 5152.21}{14 - 10} = \frac{62.18}{4} = 15.545 \][/tex]

### Interval [0, 4]
- [tex]\(t_1 = 0\)[/tex], [tex]\(B(t_1) = 5000.00\)[/tex]
- [tex]\(t_2 = 4\)[/tex], [tex]\(B(t_2) = 5060.34\)[/tex]

[tex]\[ \text{Average Rate of Change} = \frac{5060.34 - 5000.00}{4 - 0} = \frac{60.34}{4} = 15.085 \][/tex]

### Interval [0, 14]
- [tex]\(t_1 = 0\)[/tex], [tex]\(B(t_1) = 5000.00\)[/tex]
- [tex]\(t_2 = 14\)[/tex], [tex]\(B(t_2) = 5214.39\)[/tex]

[tex]\[ \text{Average Rate of Change} = \frac{5214.39 - 5000.00}{14 - 0} = \frac{214.39}{14} \approx 15.314 \][/tex]

### Interval [6, 12]
- [tex]\(t_1 = 6\)[/tex], [tex]\(B(t_1) = 5090.78\)[/tex]
- [tex]\(t_2 = 12\)[/tex], [tex]\(B(t_2) = 5183.21\)[/tex]

[tex]\[ \text{Average Rate of Change} = \frac{5183.21 - 5090.78}{12 - 6} = \frac{92.43}{6} \approx 15.405 \][/tex]

### Interval [2, 8]
- [tex]\(t_1 = 2\)[/tex], [tex]\(B(t_1) = 5030.08\)[/tex]
- [tex]\(t_2 = 8\)[/tex], [tex]\(B(t_2) = 5121.41\)[/tex]

[tex]\[ \text{Average Rate of Change} = \frac{5121.41 - 5030.08}{8 - 2} = \frac{91.33}{6} \approx 15.222 \][/tex]

Now, let's order these intervals from greatest to least average rate of change:

1. [tex]\([10, 14]\)[/tex] with an average rate of change of [tex]\(15.545\)[/tex]
2. [tex]\([6, 12]\)[/tex] with an average rate of change of [tex]\(15.405\)[/tex]
3. [tex]\([0, 14]\)[/tex] with an average rate of change of [tex]\(15.314\)[/tex]
4. [tex]\([2, 8]\)[/tex] with an average rate of change of [tex]\(15.222\)[/tex]
5. [tex]\([0, 4]\)[/tex] with an average rate of change of [tex]\(15.085\)[/tex]

Thus, the ordered list of intervals from greatest to least average rate of change is:

[tex]\[ [10, 14], [6, 12], [0, 14], [2, 8], [0, 4] \][/tex]

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